Factor 9x^3-4x

$9 x^{3} - 4 x$

Step 1

Factor $x$ out of $9 x^{3} - 4 x$ .

Factor $x$ out of $9 x^{3}$ .

$x (9 x^{2}) - 4 x$

Factor $x$ out of $- 4 x$ .

$x (9 x^{2}) + x \cdot - 4$

Factor $x$ out of $x (9 x^{2}) + x \cdot - 4$ .

$x (9 x^{2} - 4)$

Step 2

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$x ({(3 x)}^{2} - 4)$

Step 3

Rewrite $4$ as $2^{2}$ .

$x ({(3 x)}^{2} - 2^{2})$

Step 4

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 2$ .

$x ((3 x + 2) (3 x - 2))$

Remove unnecessary parentheses.

$x (3 x + 2) (3 x - 2)$